TITLE: If a linear map $T$ has a $k$-dimensional invariant subspace, does it admit an $n-k$ invariant subspace?
QUESTION [9 upvotes]: Let $V$ be an $n$-dimensional real vector space, and let $1<k<n-1$ be fixed. Let $T: V\to V$ be a linear map, and suppose that there exists a $k$-dimensional $T$ invariant subspace of $V$. 

Does there exist an $(n-k)$ $T$-invariant subspace of $V$?

The smallest possible dimensions where a counter-example might be  is when $k=2,n=5$. 
Two comments:

By duality, $T$ has a $k$-dimensional invariant subspace if and only if the dual map $T^*:V^* \to V^*$ has an $n-k$-dimensional invariant subspace. (If $U$ is $T$-invariant, then the subspace of $V^*$ whose restriction to $U$ is zero is $T^*$-invariant).
I excluded the cases $k=1$ and $k=n-1$, since I know that the answer is positive for those. Indeed, since the characteristic polynomials of $T$ and $T^{*}$ are identical, we have

$$T \, \text{  has an eigenvector  if and only if } \, T^{*} \, \text{  has an eigenvector } \tag{1}.$$
By the previous comment, we also have
$$T \, \text{  has an eigenvector  if and only if } \, T^{*} \, \text{  has a co-dimension one invariant subspace } \tag{2}.$$
Combining $(1)$ and $(2)$, we conclude that $T^*$ has an eigenvector  if and only if $T^*$ has a co-dimension one invariant subspace. Since any map is the dual of its dual, this holds for any endomorphism $T$.

REPLY [5 votes]: The result is true over any field. Let $V$ be an $n$-dimensional vector space over an arbitrary field $\mathbb{F}$, let $T \colon V \rightarrow V$ be an operator and let $U$ be a $k$-dimensional $T$-invariant subspace of $V$. Denote by
$$ U^0 := \{ \varphi \in V^{*} \, | \, \varphi(u) = 0 \,\,\,\forall u \in U \} $$
the annihilator of $U$. Note that if $\varphi \in U^{0}$ then
$$ (T^{*}(\varphi))(u) = \varphi(Tu) = 0 $$
because $Tu \in U$ and $\varphi \in U^{0}$ so $T^{*}(\varphi) \in U^{0}$ and hence $U^{0}$ is an $n-k$-dimensional $T^{*}$-invariant subspace.
Now we can use the relatively well-known but non-trivial fact that any matrix is similar to its transpose (see this answer). An invariant way of stating this is that there exists an isomorphism $S \colon V \rightarrow V^{*}$ such that $S^{-1} \circ T^{*} \circ S = T$. Set $W = S^{-1}(U^{0})$. Then $W$ is an $n-k$-dimensional subspace of $V$ and
$$ T(W) = T(S^{-1}(U^{0})) = S^{-1}(T^{*}(U^{0})) \subseteq S^{-1}(U^{0}) = W $$
so $W$ is $T$-invariant.